Exercise Construct a titration curve by plotting pH vs. VNaOH added.HCl(aq) + NaOH(aq)H₂O(1) + NaCl(aq)Volume NaOH (mL) [H'] pH0.005.0010.0015.0020.0025.0030.0035.0040.0045.0048.0050.0052.0055.0060.0065.0070.0075.0080.0085.0090951001×107Hints:The pH of the solution is calculated using the following equations:[H]=(a) Initial conditions (VNaOH = 0), [H] = [HCI](b) After initial addition of NaOH and before the end point:[H]=MHCI VHCI-MVHCI + V NaOH1x10-14XV.titrant: NaOH, 0.1M(c) At the end point (VNaOH = 50 mL), all the HCI is being consumed and the acidityof the solution comes from auto ionization of water which is:[H"]=1x107(d) After the end point:Moanalyte: HC1, 0.1 M, 50 ml.NaOH *V NaOHNaOffV +V₂C²NaOH

This is an acid-base titration. The table for graph construction is as follows

Answered: Construct a titration curve by plotting…

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Construct a titration curve by plotting pH vs. VNªOH added.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Exercise Construct a titration curve by plotting pH vs. VNaOH added. HCl(aq) + NaOH(aq) H₂O(1) + NaCl(aq) Volume NaOH (mL) [H] pH 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 48.00 50.00 $2.00 55.00 60.00 65.00 70.00 75.00 80.00 85.00 90 95 100 What is the pH when the solution volume in the flask reaches 95 ml? O a 1279 Ob 12.82 OC 1.95 O d. 12.75 De 2.28 Hints: The pH of the solution is calculated using the following equations: (a) Initial conditions (VNaOH = 0), [H"]=[HCI] (b) After initial addition of NaOH and before the end point: MC VC-M NOVNO [H"]= V LV (c) At the end point (Vou 50 mL), all the HCI is being consumed and the acidity of the solution comes from auto ionization of water which is: [H]-1-10 (d) After the end point: 1x10-¹4 M Isel Boy XV V₁W+VIRY [H"]= titrant: NaOH, 0.1M analyte: HC1, 0.1 M, 50 ml.
Exercise: Construct a titration curve by plotting pH vs. VNaOH added. HCl(aq) + NaOH(aq) H₂O(1) + NaCl(aq) Volume NaOH (mL) [H] pH 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 48.00 50.00 52.00 55.00 60.00 65.00 70.00 75.00 80.00 85.00 90 95 100 1×10² Hints: The pH of the solution is calculated using the following equations: (a) Initial conditions (VNaOH = 0), [H"]=[HCI] (b) After initial addition of NaOH and before the end point: [H]=MVMNVN V/ LV (c) At the end point (Vsaou 50 mL), all the HC1 is being consumed and the acidity of the solution comes from auto ionization of water which is: [H]-1-10 (d) After the end point: 1x 10-¹ XV VO+VIRY [H"]= NOW ABRIREDIC titrant: NaOH, 0.1M Question 1) What is the pH when the solution volume in the flask reaches 95 ml? a. 12.79 b.12.82 c. 1.95 d. 12. 75e. 2.28 Question2) What is the maximum pH reached at the end of titration? a. 12.77b. 12.79 c. 12.81 d. 12. 78 e. 12.82 analyte: HC1, 0.1 M, 50 ml.
A chemistry student is studying the effect of buffers. During his experiment, he collected the following data. However, is advisor informed him that his conclusions were incorrect. What are the likely conclusions the student provided that led to his advisor telling him this? (Select all that apply) Moles of pH of pH of buffer pH of buffer HCI water added (0.10 M HA/ (0.20 M HA/ Select 2 correct answer(s) 0.10 MA) 0.20 MA) O Buffer 2 is more concentrated, so its capacity to resist to changes in pH is higher than Buffer 1. 7.00 4.74 4.74 10-6 6.00 4.74 4.74 Any concentration of HCl will decrease the pH of water, while high concentrations of HCl is needed to change the pH of a buffer. 10-5 5.00 4.74 4.74 10-4 4.00 4.74 4.74 O Buffer 1 is less concentrated, so its capacity to resist changes in pH is higher than Buffer 2. 10-3 3.00 4.74 4.74 10-2 2.00 4.66 4.70 Adding HCl will increase the pH of water, but not the pH of the buffers 10-1 1.00 2.72 4.27
1. Practice Exercise Exactly 100 mL of 0.10 M nitrous acid (HNO,) are titrated with a 0.10 M NAOH solution. Calculate the pH for (a) the initial solution, (b) the point at which 80 mL of the base has been added, (c) the equivalence point, (d) the point at which 105 mL of the base has been added.
Exercise Construct a titration curve by plotting pH vs. VNaOH added. HCl(aq) + NaOH(aq) H₂O(1) + NaCl(aq) Volume NaOH (mL) [H*] pH 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 48.00 50.00 52.00 55.00 60.00 65.00 70.00 75.00 80.00 85.00 90 95 100 1x10¹ Hints: The pH of the solution is calculated using the following equations: (a) Initial conditions (VNaOH = 0), [H] = [HCI] (b) After initial addition of NaOH and before the end point: [H]=MCVMNam *V N₂ V LV (c) At the end point (Vsou 50 mL), all the HCI is being consumed and the acidity of the solution comes from auto ionization of water which is: [H]-1×10 (d) After the end point: [H"]= 1x10-14 xV VO+VY M NOW ɔ CLOTHING titrant: NaOH, 0.1M Question 1) What is the pH when the solution volume in the flask reaches 95 ml? a. 12.79 b. 12.82 c. 1.95 d. 12. 75e. 2.28 Question2) What is the maximum pH reached at the end of titration? a. 12.77b. 12.79 c. 12.81 d. 12.78 e. 12.82 analyte: HC1, 0.1 M, 50 ml.
Which of the following pairs of solutions will form a buffer? Select all that apply Group of answer choices 0.100 M HF and 0.100 M NaF 0.100 M HCl and 0.100 M NaCl 0.100 M NaCl and 0.100 M KCl 0.100 M HCN and 0.100 M LiCN
Which of the following can be classified as buffer solutions? (Select all that apply.) 0.50 M HOCl + 0.35 M KOCH 0.15 M HC1O4 +0.20 M RbOH 0.85 M H₂NNH2 + 0.60 M H₂NNH3 NO3 0.70 M KOH + 0.70 M HONH, 0.25 M HBr + 0.25 M HOBr ооооо
Which of the following can be classified as buffer solutions? (Select all that apply.) D 0.50 M HỌC1 + 0.35 M KOCH 0.25 M HBr + 0.25 M HOBr 0.85 M H₂NNH2 + 0.60 M H₂NNH3 NO3 D 0.70 M KOH + 0.70 M HONH2 0.15 M HC1O4 +0.20 M RbOH
Which of the following can be classified as buffer solutions? (Select all that apply.) 0.15 M HC1O4 + 0.20 M RbOH 0.85 M H₂NNH₂ + 0.60 M H₂NNH3 NO3 0.50 M HOCI + 0.35 M KOCI 0.25 M HBr + 0.25 M HOBr [ 0.70 M KOH + 0.70 M HONH,
v Question Completion Status: QUESTION 13 Consider an acid-base titration in which the base is dispenssd from a burette into a flask containing an acid. If any drops of the base adhere to the inner walls of the flask, but do not ackially mix with the solution, the calculated acid concentration would be O the same as the actual value. O lower than the actual value. O higher than the actual value. O Need more information QUESTION 14 What is the BEST numerical value for the quantity of liquid in the graduated cylinder? Save A Click Save and Submit to save and submit. Click Save All Answers to save all answers.
tation The following pH curve is for the titration of 20.0 mL of a 0.10 M solution of a weak monoprotic base with a 0.10 M solution of a strong monoprotic acid. 12.0 10.0 Data points at 1 mL increments. 8.0 Hd 6.0 4.0 2.0 0.0- 0.0 10.0 20.0 30.0 Volume of titrant (mL) 40.0 ܠܕ ܠܐ ܒ ܬ ܤ ܐ ܠܐ 50.0 CD
When a 30.0 mL sample of a 0.421 M aqueous hypochlorous acid solution is titrated with a 0.347 M aqueous sodium hydroxide solution, (1) What is the pH at the midpoint in the titration? (2) What is the pH at the equivalence point of the titration? (3) What is the pH after 54.6 mL of sodium hydroxide have been added? Submit Answer Try Another Version 3 item attempts remaining